2009 AMC 12B Problems/Problem 23

Problem

A region $S$ in the complex plane is defined by $S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.$ A complex number $z = x + iy$ is chosen uniformly at random from $S$ . What is the probability that $\left(\frac34 + \frac34i\right)z$ is also in $S$ ?

$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac23\qquad \textbf{(C)}\ \frac34\qquad \textbf{(D)}\ \frac79\qquad \textbf{(E)}\ \frac78$

Solution

We can directly compute $\left(\frac34 + \frac34i\right)z = \left(\frac34 + \frac34i\right)(x + iy) = \frac{3(x-y)}4 + \frac{3(x+y)}4 \cdot i$ .

This number is in $S$ if and only if $-1 \leq \frac{3(x-y)}4 \leq 1$ and at the same time $-1 \leq \frac{3(x+y)}4 \leq 1$ . This simplifies to $|x-y|\leq\frac 43$ and $|x+y|\leq\frac 43$ .

Let $T = \{ x + iy : |x-y|\leq\frac 43 ~\land~ |x+y|\leq\frac 43 \}$ , and let $[X]$ denote the area of the region $X$ . Then obviously the probability we seek is $\frac {[S\cap T]}{[S]} = \frac{[S\cap T]}4$ . All we need to do is to compute the area of the intersection of $S$ and $T$ . It is easiest to do this graphically:

$[asy] unitsize(2cm); defaultpen(0.8); path s = (-1,-1) -- (-1,1) -- (1,1) -- (1,-1) -- cycle; path t = (4/3,0) -- (0,4/3) -- (-4/3,0) -- (0,-4/3) -- cycle; path s_cap_t = (1/3,1) -- (1,1/3) -- (1,-1/3) -- (1/3,-1) -- (-1/3,-1) -- (-1,-1/3) -- (-1,1/3) -- (-1/3,1) -- cycle; filldraw(s, lightred, black); filldraw(t, lightgreen, black); filldraw(s_cap_t, lightyellow, black); draw( (-5/3,0) -- (5/3,0), dashed ); draw( (0,-5/3) -- (0,5/3), dashed ); [/asy]$

Coordinate axes are dashed, $S$ is shown in red, $T$ in green and their intersection is yellow. The intersections of the boundary of $S$ and $T$ are obviously at $(\pm 1,\pm 1/3)$ and at $(\pm 1/3,\pm 1)$ .

Hence each of the four red triangles is an isosceles right triangle with legs long $\frac 23$ , and hence the area of a single red triangle is $\frac 12 \cdot \left( \frac 23 \right)^2 = \frac 29$ . Then the area of all four is $\frac 89$ , and therefore the area of $S\cap T$ is $4 - \frac 89$ . Then the probability we seek is $\frac{ [S\cap T]}4 = \frac{ 4 - \frac 89 }4 = 1 - \frac 29 = \boxed{\frac 79}$ .

(Alternately, when we got to the point that we know that a single red triangle is $\frac 29$ , we can directly note that the picture is symmetric, hence we can just consider the first quadrant and there the probability is $1 - \frac 29 = \frac 79$ . This saves us the work of first multiplying and then dividing by $4$ .)

Solution 2 (Same idea)

The solution proposed above is good, but there is a more straightforward method. First, turn $\frac34 + \frac34i$ into polar form as $\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}$ . Restated using geometric probabilities, we are trying to find the portion of a square enlarged by a factor of $\frac{3\sqrt{2}}{4}$ and rotated $45\degree$ (Error compiling LaTeX. Unknown error_msg)

$[asy] unitsize(2cm); defaultpen(0.8); path s = (-1,-1) -- (-1,1) -- (1,1) -- (1,-1) -- cycle; path t = (4/3,0) -- (0,4/3) -- (-4/3,0) -- (0,-4/3) -- cycle; path s_cap_t = (1/3,1) -- (1,1/3) -- (1,-1/3) -- (1/3,-1) -- (-1/3,-1) -- (-1,-1/3) -- (-1,1/3) -- (-1/3,1) -- cycle; filldraw(s, lightred, black); filldraw(t, lightgreen, black); filldraw(s_cap_t, lightyellow, black); draw( (-5/3,0) -- (5/3,0), dashed ); draw( (0,-5/3) -- (0,5/3), dashed ); [/asy]$

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2009 AMC 12B Problems/Problem 23

Contents

Problem

Solution

Solution 2 (Same idea)

See Also